Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> adj[100005], id[100005];
int sp[100005][21], lvl[100005] = {0};
int n, m, q;
vector<int> lst[100005][21];
void dfs(int r, int p) {
  sp[r][0] = p ? p : -1;
  lvl[r] = lvl[p] + 1;
  sort(id[r].begin(), id[r].end());
  for (int i = 0; i < min(10, (int)id[r].size()); i++) {
    lst[r][0].push_back(id[r][i]);
  }
  for (auto i : adj[r]) {
    if (i == p) continue;
    dfs(i, r);
  }
}
void preprocess() {
  for (int j = 1; j <= 20; j++) {
    for (int i = 1; i <= n; i++) {
      if (sp[i][j - 1] != -1) {
        sp[i][j] = sp[sp[i][j - 1]][j - 1];
      }
    }
  }
  for (int j = 1; j <= 20; j++) {
    for (int i = 1; i <= n; i++) {
      if (sp[i][j - 1] == -1) continue;
      lst[i][j].resize(lst[i][j - 1].size() + lst[sp[i][j - 1]][j - 1].size());
      merge(lst[i][j - 1].begin(), lst[i][j - 1].end(),
            lst[sp[i][j - 1]][j - 1].begin(), lst[sp[i][j - 1]][j - 1].end(),
            lst[i][j].begin());
      lst[i][j].resize(min(10, (int)lst[i][j].size()));
    }
  }
}
void Find(int x, int dis, vector<int> &ans) {
  for (int i = 20; i >= 0; i--) {
    if (dis >= (1ll << i)) {
      dis -= (1ll << i);
      for (auto j : lst[x][i]) ans.push_back(j);
      x = sp[x][i];
    }
  }
}
int LCA(int u, int v) {
  if (lvl[u] < lvl[v]) {
    swap(u, v);
  }
  for (int i = 20; i >= 0; i--) {
    if (sp[u][i] != -1 && lvl[sp[u][i]] >= lvl[v]) {
      u = sp[u][i];
    }
  }
  if (u == v) return u;
  for (int i = 20; i >= 0; i--) {
    if (sp[u][i] != -1 && sp[u][i] != sp[v][i]) {
      u = sp[u][i];
      v = sp[v][i];
    }
  }
  return sp[v][0];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(sp, -1, sizeof sp);
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    int tem;
    cin >> tem;
    id[tem].push_back(i);
  }
  dfs(1, 0);
  preprocess();
  for (int i = 1; i <= q; i++) {
    int x, y, k;
    cin >> x >> y >> k;
    vector<int> ans;
    int lca = LCA(x, y);
    Find(x, lvl[x] - lvl[lca] + 1, ans);
    Find(y, lvl[y] - lvl[lca], ans);
    sort(ans.begin(), ans.end());
    ans.resize(min((int)ans.size(), k));
    cout << ans.size() << " ";
    for (auto i : ans) {
      cout << i << " ";
    }
    cout << endl;
  }
  return 0;
}